Proof. Let $\( \mathrm{B}_{X}\mathopen{}\left( r, {x}_{0}\right)\mathclose{} \)$ denote the open ball of radius $\( r \)$ around $\( {x}_{0} \)$ in $\( X \)$, and likewise for balls in $\( Y \)$. Write $\( \mathrm{B}\mathopen{}\left( r\right)\mathclose{}= \mathrm{B}_{X}\mathopen{}\left( r, 0\right)\mathclose{} \)$. It will suffice to show that $\( T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} \)$ contains a ball about $\( 0 \)$ in $\( Y \)$. Since $\( X= \bigcup_{n=1}^{\infty}{} \mathrm{B}\mathopen{}\left( n\right)\mathclose{} \)$ and $\( T \)$ is surjective, we have that $\( Y= \bigcup_{n=1}^{\infty}{} T\mathopen{}\left( \mathrm{B}\mathopen{}\left( n\right)\mathclose{}\right)\mathclose{} \)$. The map $\( y\mapsto ny \)$ is a homeomorphism of $\( Y \)$ that maps $\( T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} \)$ to $\( T\mathopen{}\left( \mathrm{B}\mathopen{}\left( n\right)\mathclose{}\right)\mathclose{} \)$, so it follows from Baire's Theorem that $\( \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$ contains an open ball about $\( 0 \)$. Thus there exists $\( {y}_{0}\in Y \)$ and an $\( r\gt 0 \)$ such that $\( \mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$. Now pick $\( {y}_{1}= T\mathopen{}\left( x\right)\mathclose{} \)$ with $\( x\in \mathrm{B}\mathopen{}\left( 1\right)\mathclose{} \)$ such that $\( \mathopen{}\left\lVert{}{y}_{1}-{y}_{0}\right\rVert\mathclose{}\lt 2r \)$. Then $\( \mathrm{B}_{Y}\mathopen{}\left( 2r, {y}_{1}\right)\mathclose{}\subseteq \mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$. If $\( \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt 2r \)$, then $\( {y}_{1}-y\in \mathrm{B}_{Y}\mathopen{}\left( 2r, {y}_{1}\right)\mathclose{}\subseteq \mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$, and hence $\( {y}_{1}-y\in \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$. Then $$\[ y= {y}_{1}+\mathopen{}\left(y-{y}_{1}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}+\mathopen{}\left(y-{y}_{1}\right)\mathclose{}\in T\mathopen{}\left( x\right)\mathclose{}+\overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }= \overline{ T\mathopen{}\left( \mathopen{}\left({x}_{1}+\mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} }\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 2\right)\mathclose{}\right)\mathclose{} } \]$$. So what we have is that $\( \mathrm{B}_{Y}\mathopen{}\left( 2r, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 2\right)\mathclose{}\right)\mathclose{} } \)$, and hence $\( \mathrm{B}_{Y}\mathopen{}\left( r, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$.

We will conclude the proof by showing that $\( \mathrm{B}_{Y}\mathopen{}\left( \frac{r}{2}, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$. Now, since $\( T \)$ commutes with dilations, we have that $\( \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt {2}^{{-}n}r \)$ implies $\( y\in \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( {2}^{{-}n} \right)\mathclose{}\right)\mathclose{} } \)$. Suppose that $\( \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt \frac{r}{2} \)$. We can find $\( {x}_{1}\in \mathrm{B}_{X}\mathopen{}\left( \frac{1}{2}, 0\right)\mathclose{} \)$ such that $\( \mathopen{}\left\lVert{}y-T\mathopen{}\left( {x}_{1}\right)\mathclose{}\right\rVert\mathclose{}\lt \frac{r}{4} \)$. By induction we get $\( {x}_{n}\in \mathrm{B}\mathopen{}\left( {2}^{{-}n} \right)\mathclose{} \)$ such that $\( \mathopen{}\left\lVert{}y-\sum_{i=1}^{n}{} T\mathopen{}\left( {x}_{i}\right)\mathclose{} \right\rVert\mathclose{}\lt r{2}^{{-}n-1} \)$. Since $\( X \)$ is complete the series $\( \sum_{n=1}^{\infty}{} {x}_{n} \)$ converges in $\( X \)$. Call the sum $\( s \)$. Then $\( \mathopen{}\left\lVert{}s\right\rVert\mathclose{}\lt \sum_{n=1}^{\infty}{} {2}^{{-}n} = 1 \)$ and $\( y= T\mathopen{}\left( s\right)\mathclose{} \)$ (because $\( T \)$ is bounded). Thus $\( y\in T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} \)$ for all $\( y \)$ such that $\( \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt \frac{r}{2} \)$.

Proof. Define $\( {φ}_{1}\in \mathcal{L}\mathopen{}\left( \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}, X\right)\mathclose{} \)$ by $\( {φ}_{1}\mathopen{}\left( x, T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= x \)$ and $\( {φ}_{2}\in \mathcal{L}\mathopen{}\left( \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}, Y\right)\mathclose{} \)$ by $\( {φ}_{2}\mathopen{}\left( x, T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{} \)$. It is easily seen that $\( X\oplus Y \)$ is complete, so our hypothesis makes $\( \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{} \)$ complete. Notice that $\( {φ}_{1} \)$ is bijective. By Corollary II.18, $\( {φ}_{1}^{-1} \)$ is bounded. Thus $\( {φ}_{2}\circ {φ}_{1}^{-1}= T \)$ is bounded.